A line makes angles `alpha,beta,gamma`and `delta` with the diagonals of a cube, then find the value of `cos^2alpha+cos^2beta+cos^2gamma+cos^2delta`.

Options

A. `1/3`
B. `2/3`
C. `1`
D. `4/3`

Explanation

To solve the problem of finding the value of \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta \) where a line makes angles \( \alpha, \beta, \gamma, \) and \( \delta \) with the diagonals of a cube, we can follow these steps:

### Step 1: Identify the Diagonals of the Cube
A cube has four main diagonals. Let's denote them as:
- Diagonal AL
- Diagonal BM
- Diagonal CN
- Diagonal OP

### Step 2: Determine the Direction Cosines of the Diagonals
For a cube with side length \( a \), the direction cosines of the diagonals can be calculated. The direction cosines for each diagonal are:
- For diagonal OP: \( \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \)
- For diagonal AL: \( \left( -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \)
- For diagonal BM: \( \left( \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \)
- For diagonal CN: \( \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right) \)

### Step 3: Define the Direction Cosines of the Line
Let the direction cosines of the line be \( L, M, N \). The angles \( \alpha, \beta, \gamma, \delta \) are the angles made by the line with the diagonals.

### Step 4: Write the Cosine Relations
Using the direction cosines, we can express the cosines of the angles:
1. \( \cos \alpha = \frac{L + M + N}{\sqrt{3}} \)
2. \( \cos \beta = \frac{-L + M + N}{\sqrt{3}} \)
3. \( \cos \gamma = \frac{L - M + N}{\sqrt{3}} \)
4. \( \cos \delta = \frac{L + M - N}{\sqrt{3}} \)

### Step 5: Square Each Cosine
Now, we will square each of these equations:
1. \( \cos^2 \alpha = \frac{(L + M + N)^2}{3} \)
2. \( \cos^2 \beta = \frac{(-L + M + N)^2}{3} \)
3. \( \cos^2 \gamma = \frac{(L - M + N)^2}{3} \)
4. \( \cos^2 \delta = \frac{(L + M - N)^2}{3} \)

### Step 6: Add the Squared Cosines
Now, we sum these squared values:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{1}{3} \left( (L + M + N)^2 + (-L + M + N)^2 + (L - M + N)^2 + (L + M - N)^2 \right) \]

### Step 7: Simplify the Expression
Expanding each term:
1. \( (L + M + N)^2 = L^2 + M^2 + N^2 + 2LM + 2LN + 2MN \)
2. \( (-L + M + N)^2 = L^2 + M^2 + N^2 - 2LM + 2LN + 2MN \)
3. \( (L - M + N)^2 = L^2 + M^2 + N^2 - 2LM + 2LN - 2MN \)
4. \( (L + M - N)^2 = L^2 + M^2 + N^2 + 2LM - 2LN - 2MN \)

Adding these gives:
\[ 4(L^2 + M^2 + N^2) \]

### Step 8: Substitute the Value of \( L^2 + M^2 + N^2 \)
Since the direction cosines satisfy \( L^2 + M^2 + N^2 = 1 \):
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{4}{3} (L^2 + M^2 + N^2) = \frac{4}{3} \]

### Conclusion
Thus, we have shown that:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{4}{3} \]